Consider the two balls, A and B. In the question, the initial speed of ball B placed at the rest position is zero. Use the momentum conservation relation to find the velocity and mass of the target ball.

Given data

The mass of the ball is\({m_{\rm{A}}} = 0.220\;{\rm{kg}}\).

The speed of the ball is\({v_{\rm{A}}} = 5.5\;{\rm{m/s}}\).

The rebound speed of the ball is\({v'_{\rm{A}}} = - 3.8\;{\rm{m/s}}\).

The relation of relative velocity is given by the following equation:

\(\begin{array}{c}{v_{\rm{A}}} - {v_{\rm{B}}} = - \left( {{{v'}_{\rm{A}}} - {{v'}_{\rm{B}}}} \right)\\{v_{\rm{A}}} - 0 = - \left( {{{v'}_{\rm{A}}} - {{v'}_{\rm{B}}}} \right)\\{{v'}_{\rm{B}}} = {v_{\rm{A}}} + {{v'}_{\rm{A}}}\end{array}\)

Here,\({v'_{\rm{B}}}\)is the final speed of the target ball, and\({v_{\rm{B}}}\)is the initial velocity of the second ball, whose value is zero.

On plugging the values in the above equation,

\(\begin{array}{c}{{v'}_{\rm{B}}} = \left( {5.5\;{\rm{m/s}}} \right) + \left( { - 3.8\;{\rm{m/s}}} \right)\\{{v'}_{\rm{B}}} = 1.7\;{\rm{m/s}}\end{array}\).

Thus, \({v'_{\rm{B}}} = 1.7\;{\rm{m/s}}\) is the required speed.

The relation from the momentum equation is as follows:

\(\begin{array}{c}P = P'\\{m_{\rm{A}}}{v_{\rm{A}}} + {m_{\rm{B}}}{v_{\rm{B}}} = {m_{\rm{A}}}{{v'}_{\rm{A}}} + {m_{\rm{B}}}{{v'}_{\rm{B}}}\end{array}\)

Here, \({m_{\rm{B}}}\) is the mass of the target ball.

On plugging the values in the above equation,

\(\begin{array}{c}\left( {0.220\;{\rm{kg}}} \right)\left( {5.5\;{\rm{m/s}}} \right) + 0 = \left( {0.220\;{\rm{kg}}} \right)\left( { - 3.8\;{\rm{m/s}}} \right) + {m_{\rm{B}}}\left( {1.7\;{\rm{m/s}}} \right)\\{m_{\rm{B}}} = 1.20\;{\rm{kg}}\end{array}\).

Thus, \({m_{\rm{B}}} = 1.20\;{\rm{kg}}\) is the required mass of the target ball.